Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Full CI computing power

  1. Aug 12, 2013 #1
    For a molecule of [itex]n[/itex] nuclei and [itex]x[/itex] total electrons, how can I work out how much computing power or processing time is required for an exact Full CI calculation for that molecule?
     
  2. jcsd
  3. Aug 13, 2013 #2

    DrDu

    User Avatar
    Science Advisor

    For an exact full CI you need a complete MO basis which contains an infinity of molecular orbitals and you can form an infinity of Slater determinants from it. Hence you will need infinite computing power for more than one electron in general.
    So you have to be less ambitious. E.g. you can chose a certain finite basis set, e.g. VQZ. Let's say you have 100 basis functions per atom (quite a big basis already) (this will also depend on the type of atom). If you have 10 atoms with 10 electrons each, you can form 100 out of 1000 Slater determinants (roughly 10^300) so this will still take almost forever.
     
  4. Aug 13, 2013 #3
    Hang on, please clarify. Ok so as you've pointed out an exact solution will take infinite time, i.e. impossible, except for one-electron systems. If you have a certain basis set with k basis functions per nucleus, and n nuclei with x total electrons ... I still don't get how you are proposing to work out the amount of computing power needed? Thanks for the guidance. Can you do an example, maybe, let there be k basis functions, but now we're talking about an H2O molecule (n=3, x=10), how much processing power/time is needed?
     
  5. Aug 14, 2013 #4

    DrDu

    User Avatar
    Science Advisor

    So if you have k basis functions and x electrons, you have 2k spin orbitals. As each Slater determinant has to be made up from x different spin orbitals, you can form $$m=\begin{pmatrix} 2k\\x \end{pmatrix}\approx x^{2k}$$ different Slater determinants (you could classify them by spin and symmetry to reduce the number, but we are interested in the gross picture). So basically your Hamiltonian becomes an $$m \times m$$ matrix which you have to diagonalize. The matrix will be sparse as the two electron matrix elements vanish between most of the Slater determinants but still the size increases factorially with the size of the system.
     
  6. Aug 14, 2013 #5
    Ok thanks.
    So what value of k would we typically use for something very simple like a He molecule or H- ion?

    And with the same k, let's say processing time on a certain computer is 10 seconds for a H- ion (2 electrons); will it then be 10!/2!*10 seconds for a (10-electron) water molecule?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Full CI computing power
  1. Computer programs? (Replies: 3)

Loading...